# Set theory question

• Nov 2nd 2005, 12:06 PM
juef
Set theory question
Hi all,

As a homework, I've been asked to find whether there is or not an infinite group such that all of its proper subgroups are finite.

I've been told to consider {m/2^n, m,n integers, 0<=m/2^n<1} with mod1 addition. It's not very hard to show that it's an infinite group, but I can't see how (for example) G\{1/2} could be a finite group, let alone showing that all of G's subgroups are finite...

Can anyone help me? Thank you! :)
• Nov 22nd 2005, 09:06 PM
BubbleBrain_103
G has no proper infinite subgroups
I take it that your group G is {m/2^n, m,n integers, 0<=m/2^n<1}, the group of diadic rationals mod 1.

when you write G/{1/2}, I take you to mean the quotient (or factor) group of G with respect to the subgroup generated by the element 1/2 in G. This certainly is an infinite group (as any infinite group mod a finite subgroup must be). However, G/{1/2} is not a subgroup of G, it is a factor group of a subgroup of G, and so does not provide a counterexample.

We can show that all of G's proper subgroups are finite as follows.

Let H be an infinite subgroup of G -- we wish to show then the H must in fact be equal to all of G. First note that if m/2^n is an element of G, then it is a multiple of any element of the form 1/2^(n+k), where k is any non-negative integer [i.e., m/2^n = (m*2^k)*(1/2^(n+k)) ]. Thus if we can show that for any integers m and n, there is an integer k >= n such that 1/2^k is in H, we will have shown that m/2^n is also in H; and this in turn shows that H must contain every element of G, and hence H = G.

The first thing to notice is that for any n, there are only finitely many elements of G of the form m/2^n (if we abide by the convention that m/2^n is in lowest terms, i.e., assume m is an odd number). Then since H is infinite, there cannot be a bound on the powers of 2 occuring in the denominators of the elements of H. Thus for arbitrarily large n, we can find an element of the form m/2^n in H, with m odd.

The second thing to notice is that if m/2^n is in lowest terms, then it is of order 2^n, since 2^n is the smallest integer k that satisfies k*(m/2^n) = 0 mod 1 (i.e., that makes k*(m/2^n) an integer). But notice also that the subgroup generated by 1/2^n is also of order 2^n. Thus since {1/2^n} is a subgroup of which m/2^n is a member, and since m/2^n generates a subgroup of equal size, we have that {1/2^n} = {m/2^n}. More importantly, we have that 1/2^n is a member of {m/2^n}. And since {m/2^n} is a subgroup of H, 1/2^n must be a member of H.

Thus we know that there are elements in H that are of the form 1/2^n, for arbitrarily large n, and from what was said previously, H = G. Thus G has no infinite proper subgroups.

Let me know if that helped.
• Nov 23rd 2005, 07:00 AM
juef
Wow, thank you very much for the detailed explanation! :) It certainly helps me. I hadn't thought trying to show it that way... but it sure works, eh? :D

Thank you very much!